# Integral Test

## Contents

## Theorem

Let $f$ be a real function which is continuous, positive and decreasing on the interval $\left[{1 \,.\,.\, +\infty}\right)$.

Let the sequence $\left \langle {\Delta_n} \right \rangle$ be defined as:

- $\displaystyle \Delta_n = \sum_{k \mathop = 1}^n f \left({k}\right) - \int_1^n f \left({x}\right) \rd x$

Then $\left \langle {\Delta_n} \right \rangle$ is decreasing and bounded below by zero.

Hence it converges.

## Proof

From Upper and Lower Bounds of Integral, we have that:

- $\displaystyle m \left({b - a}\right) \le \int_a^b f \left({x}\right) \rd x \le M \left({b - a}\right)$

where:

- $M$ is the maximum

and:

- $m$ is the minimum

of $f \left({x}\right)$ on $\left[{a \,.\,.\, b}\right]$.

Since $f$ decreases, $M = f \left({a}\right)$ and $m = f \left({b}\right)$.

Thus it follows that:

- $\displaystyle \forall k \in \N_{>0}: f \left({k + 1}\right) \le \int_k^{k + 1} f \left({x}\right) \rd x \le f \left({k}\right)$

as $\left({k + 1}\right) - k = 1$.

Thus:

\(\displaystyle \Delta_{n + 1} - \Delta_n\) | \(=\) | \(\displaystyle \left({\sum_{k \mathop = 1}^{n + 1} f \left({k}\right) - \int_1^{n + 1} f \left({x}\right) \rd x}\right) - \left({\sum_{k \mathop = 1}^n f \left({k}\right) - \int_1^n f \left({x}\right) \rd x}\right)\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle f \left({n + 1}\right) - \int_n^{n + 1} f \left({x}\right) \rd x\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle f \left({n + 1}\right) - f \left({n + 1}\right)\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 0\) | $\quad$ | $\quad$ |

Thus $\left \langle {\Delta_n} \right \rangle$ is decreasing.

$\Box$

Also:

\(\displaystyle \Delta_n\) | \(=\) | \(\displaystyle \sum_{k \mathop = 1}^n f \left({k}\right) - \sum_{k \mathop = 1}^{n-1} \int_k^{k+1} f \left({x}\right) \ \mathrm dx\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(\ge\) | \(\displaystyle \sum_{k \mathop = 1}^n f \left({k}\right) - \sum_{k \mathop = 1}^{n-1} f \left({k}\right)\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle f \left({n}\right)\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(\ge\) | \(\displaystyle 0\) | $\quad$ | $\quad$ |

$\Box$

Hence the result.

$\blacksquare$

## Also known as

The **integral test** is also known as the **Euler-Maclaurin Summation Formula**, but that result properly refer to a more precise theorem of which this is a simple corollary.

## Notes

It follows from this that if $f$ is continuous, positive and decreasing on $\left[{1 \,.\,.\, \infty}\right)$, then the series $\displaystyle \sum_{k \mathop = 1}^\infty f \left({k}\right)$ and the improper integral $\displaystyle \int_1^{\mathop \to +\infty} f \left({x}\right) \rd x$ either both converge or both diverge.

So this theorem provides a test for the convergence of both a series and an improper integral.

## Sources

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 13.32$ - 1992: Larry C. Andrews:
*Special Functions of Mathematics for Engineers*... (previous) ... (next): $\S 1.2.2$: Summary of convergence tests: Theorem $1.2$